i am learning how to solve quadratic inequalities graphically but it is the same steps every single time  i dont want to learn algorithms i want to learn maths so my doubt is
1) Why should the inequality be written with 0 alone on one side like ax^2+bx+c<0
2) What is the reason that makes the x axis taken to be that 0

tryingtomaths
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 29092016 15:06

old_engineer
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 29092016 15:31
1) Solving for ax^2+bx+c=0 will give you the socalled critical values of x, in other words the values of x at which the expression transitions between being <0 and >0.
2) y=0 is just another way of identifying the x axis. 
Notnek
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 29092016 15:41
(Original post by tryingtomaths)
i am learning how to solve quadratic inequalities graphically but it is the same steps every single time  i dont want to learn algorithms i want to learn maths so my doubt is
1) Why should the inequality be written with 0 alone on one side like ax^2+bx+c<0
2) What is the reason that makes the x axis taken to be that 0
Let's say you're solving the inequality
Then consider the function below
Since is equal to , if then this gives us the original inequality
So solving is equivalent to finding where is in the graph of .
To find this, you first need to find where to give you the so called critical values where y switches from positive to negative (or viceversa).
The reason why you start by rearranging into the form is because you'll at some point need to solve the corresponding quadratic equation as shown above. And to do this you have to put all the terms on one side. You will have learnt this at GCSE.Last edited by Notnek; 29092016 at 15:43. 
tryingtomaths
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 29092016 16:07
(Original post by notnek)
If you've been taught this as a set of steps with no explanation then you're being taught badly.
Let's say you're solving the inequality
Then consider the function below
Since is equal to , if then this gives us the original inequality
So solving is equivalent to finding where is in the graph of .
To find this, you first need to find where to give you the so called critical values where y switches from positive to negative (or viceversa).
The reason why you start by rearranging into the form is because you'll at some point need to solve the corresponding quadratic equation as shown above. And to do this you have to put all the terms on one side. You will have learnt this at GCSE.Last edited by tryingtomaths; 29092016 at 16:15. 
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 29092016 16:51
(Original post by tryingtomaths)
Thank you so much for your beautiful explanation; yes so far i can follow up, but i still don't understand why we linked the "y" from the function to replace the original question; yes i realize they are equal but so did the equation when we made it equal to 0; yet we did'nt say that 0< 0; what made us choose to say that y < 0? sorry if my question is stupid
I am not actually saying that solving is the same as solving .
You're right : if both of these are true : and then you could derive , which doesn't make sense.
If you're unsure of the above, try to ignore it for now and focus on the following:
What I am saying is that solving
is equivalent to solving this pair
and .
The reason these are equivalent is because writing means that is the same as .
So if and is the same as then it must be true that
If you're still unsure, please let me know the exact line of my post where you first were confused. 
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 29092016 17:24
(Original post by notnek)
Firstly, solving is part of the method for solving the inequality.
I am not actually saying that solving is the same as solving .
You're right : if both of these are true : and then you could derive , which doesn't make sense.
If you're unsure of the above, try to ignore it for now and focus on the following:
What I am saying is that solving
is equivalent to solving this pair
and .
The reason these are equivalent is because writing means that is the same as .
So if and is the same as then it must be true that
If you're still unsure, please let me know the exact line of my post where you first were confused. 
Notnek
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 29092016 17:28
(Original post by tryingtomaths)
what i still don't understand and can't wrap my head around is why and how we derived that when y=x^2+3x4; y moves up and we find the inequality with respect to it; however when we put some other number the same rule doesn't apply. ; i even read the purplemaths article on quadratic inequalities and understood it well but still stuck at that same problem.Thank you so much for your patience with me.
Was there any part of my last post that didn't make sense? 
tryingtomaths
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 29092016 17:30
(Original post by notnek)
I'm not sure what you mean by "y moves up". Can you explain this with an example?
Was there any part of my last post that didn't make sense?
And yes

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 29092016 17:34
It would be useful if you start with x^2 + 3x  4 < 0 and go through the method of solving it (post all the working in full) and then stop at the exact point where you beome unsure of what's going on.
I recommend you look for videos on this topic. Hopefully someone else explaining it visually will help.Last edited by Notnek; 29092016 at 17:35. 
tryingtomaths
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 29092016 17:49
(Original post by notnek)
Ignore that sentence you quoted  it's not important. The part below that sentence is where I've tried to explain your issue. Can you tell me which line make sense?
It would be useful if you start with x^2 + 3x  4 < 0 and go through the method of solving it (post all the working in full) and then stop at the exact point where you beome unsure of what's going on.
I recommend you look for videos on this topic. Hopefully someone else explaining it visually will help. x^{2}+3x4<0
 (x+4)(x1)<0
 y=(x+4)(x1)
 0=(x+4)(x1)
 x=4 l x=1
 since y<0
 4<x<1
Last edited by tryingtomaths; 29092016 at 17:58. 
Notnek
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 29092016 18:01
(Original post by tryingtomaths)
Very well; x^{2}+3x4<0
 (x+4)(x1)<0
 y=(x+4)(x1)
 0=(x+4)(x1)
 x=4 l x=1
I'm going to repost my explanation of what I think is your question (let me know if it isn't). Please tell me if there's anything that doesn't make sense.
What I am saying is that solving
is equivalent to solving this pair
and .
The reason these are equivalent is because writing means that is the same as .
So if and is the same as then it must be true that
I would be grateful if someone else reading this thread can attempt to explain. 
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 29092016 18:08
(Original post by notnek)
Thank you for that. Since you've watched all the videos without understanding, an explanation on an internet forum may not be enough. It would benefit you to have a face to face explanation of this where the you can go back and forth with the teacher until you understand.
I'm going to repost my explanation of what I think is your question (let me know if it isn't). Please tell me if there's anything that doesn't make sense.
What I am saying is that solving
is equivalent to solving this pair
and .
The reason these are equivalent is because writing means that is the same as .
So if and is the same as then it must be true that
I would be grateful if someone else reading this thread can attempt to explain.Last edited by tryingtomaths; 29092016 at 18:09. 
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 29092016 18:39
(Original post by tryingtomaths)
yes this is exactly my question and the explanation makes senses; and because it make's sense i am asking why can't we say 0<0 if we have x^2+3x4=0 in our steps? i am really not stupid i just question alot
We're not acutually saying that x^2+3x4=0. It's just part of the method.
You're not stupid. Actually you're the kind of student I like to teach  someone who asks questions. 
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 29092016 19:02
(Original post by notnek)
Solving x^2+3x4=0 is useful so that we can draw y = x^2 + 3x  4 and then use the graph to solve x^2+3x4<0
We're not acutually saying that x^2+3x4=0. It's just part of the method.
You're not stupid. Actually you're the kind of student I like to teach  someone who asks questions. 
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 29092016 19:07
(Original post by tryingtomaths)
I think i got it. So basically the function doesn't =0; we equate it to 0 just to find the points of intersection at y=0, while the whole curve will be equal to y. and by saying 0<0 we are only considering those points at the x axis rather than the whole parabola. Thank you i got it now; it's been 2 days i am trying to understand it. and thank you so much for your kind comment. you made my day
Make sure you continue to ask questions on here if there's something like this that wasn't clear to you. There are plenty of members here who will take a lot of time to explain a concept and won't stop until you're happy 
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 29092016 19:15
(Original post by notnek)
I'm not completely sure what you mean when you talk about "0<0" but if it all makes sense to you then that's what matters.
Make sure you continue to ask questions on here if there's something like this that wasn't clear to you. There are plenty of members here who will take a lot of time to explain a concept and won't stop until you're happy
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